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If \(x\) is complex, the expression \(\frac{x^2+34 x-71}{x^2+2 x-7}\) takes all values which lie in the interval \((a, b)\), find the value of \(a\) and \(b\).
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Verified Answer
The correct answer is:
\(a=5, b=9\)
Here we have to find range of
\(\frac{x^2+34 x-71}{x^2+2 x-7}=y\) (Let)
Then,
\(\begin{gathered}
x^2+34 x-71=x^2 y+2 x y-7 y \\
\Rightarrow \quad x^2(y-1)+x(2 y-34)+(71-7 y)=0
\end{gathered}\)
As \(x\) is complex, directriminant of above equation,
\(\begin{array}{ll}
& D \leq 0 \\
\Rightarrow & (2 y-34)^2-4(y-1)(71-7 y) \leq 0 \\
\Rightarrow & 5 < y < 9 \\
\Rightarrow & (a, b) \equiv(5,9)
\end{array}\)
\(\frac{x^2+34 x-71}{x^2+2 x-7}=y\) (Let)
Then,
\(\begin{gathered}
x^2+34 x-71=x^2 y+2 x y-7 y \\
\Rightarrow \quad x^2(y-1)+x(2 y-34)+(71-7 y)=0
\end{gathered}\)
As \(x\) is complex, directriminant of above equation,
\(\begin{array}{ll}
& D \leq 0 \\
\Rightarrow & (2 y-34)^2-4(y-1)(71-7 y) \leq 0 \\
\Rightarrow & 5 < y < 9 \\
\Rightarrow & (a, b) \equiv(5,9)
\end{array}\)
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